Question: Solve for $k$, $ -\dfrac{k + 1}{25k^3} = -\dfrac{3}{5k^3} - \dfrac{8}{20k^3} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25k^3$ $5k^3$ and $20k^3$ The common denominator is $100k^3$ To get $100k^3$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{k + 1}{25k^3} \times \dfrac{4}{4} = -\dfrac{4k + 4}{100k^3} $ To get $100k^3$ in the denominator of the second term, multiply it by $\frac{20}{20}$ $ -\dfrac{3}{5k^3} \times \dfrac{20}{20} = -\dfrac{60}{100k^3} $ To get $100k^3$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{8}{20k^3} \times \dfrac{5}{5} = -\dfrac{40}{100k^3} $ This give us: $ -\dfrac{4k + 4}{100k^3} = -\dfrac{60}{100k^3} - \dfrac{40}{100k^3} $ If we multiply both sides of the equation by $100k^3$ , we get: $ -4k - 4 = -60 - 40$ $ -4k - 4 = -100$ $ -4k = -96 $ $ k = 24$